LeetCode in Kotlin

576. Out of Boundary Paths

Medium

There is an m x n grid with a ball. The ball is initially at the position [startRow, startColumn]. You are allowed to move the ball to one of the four adjacent cells in the grid (possibly out of the grid crossing the grid boundary). You can apply at most maxMove moves to the ball.

Given the five integers m, n, maxMove, startRow, startColumn, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it modulo 109 + 7.

Example 1:

Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0

Output: 6

Example 2:

Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1

Output: 12

Constraints:

• 1 <= m, n <= 50
• 0 <= maxMove <= 50
• 0 <= startRow < m
• 0 <= startColumn < n

Solution

class Solution {
    private val dRowCol = arrayOf(intArrayOf(1, 0), intArrayOf(-1, 0), intArrayOf(0, 1), intArrayOf(0, -1))
    private fun dfs(
        m: Int,
        n: Int,
        remainingMoves: Int,
        currRow: Int,
        currCol: Int,
        cache: Array<Array<IntArray>>,
    ): Int {
        if (currRow < 0 || currRow == m || currCol < 0 || currCol == n) {
            return 1
        }
        if (remainingMoves == 0) {
            return 0
        }
        if (cache[currRow][currCol][remainingMoves] == -1) {
            var paths = 0
            for (i in 0..3) {
                val newRow = currRow + dRowCol[i][0]
                val newCol = currCol + dRowCol[i][1]
                val m1 = 1000000007
                paths = (paths + dfs(m, n, remainingMoves - 1, newRow, newCol, cache)) % m1
            }
            cache[currRow][currCol][remainingMoves] = paths
        }
        return cache[currRow][currCol][remainingMoves]
    }

    fun findPaths(m: Int, n: Int, maxMoves: Int, startRow: Int, startCol: Int): Int {
        val cache = Array(m) {
            Array(n) {
                IntArray(
                    maxMoves + 1,
                )
            }
        }
        for (c1 in cache) {
            for (c2 in c1) {
                c2.fill(-1)
            }
        }
        return dfs(m, n, maxMoves, startRow, startCol, cache)
    }
}

This site is open source. Improve this page.